# Exercise 4, section 2.2 of Hoffman's Linear Algebra

The problem is:

Let $$W$$ be the set of all $$(x_1,x_2,x_3,x_4,x_5)$$ in $$\mathbb{R}^5$$ which satisfy

$$\begin{array}2x_1-x_2+\frac43x_3-x_4=0\\ x_1+\frac23x_3-x_5=0\\ 9x_1-3x_2+6x_3-3x_4-3x_5=0\end{array}$$

Find a finite set of vectors which spans $$W$$

The calculation shows its simplifies to row echelon form

$$\begin{pmatrix}1&0&\frac23&0&-1\\0&1&0&1&-2\\0&0&0&0&0\end{pmatrix}$$

And the solution says:

Thus, the system is equivalent to: $$x_1+\frac23x_3-x_5=0\\ x_2+x_4-2x_5=0$$

And it proceeds by saying that:

Thus the system is parametrized by $$(x_3,x_4,x_5)$$

Setting each equal to $$1$$ and the other $$2$$ equal to $$0$$, in turn, gives the three vectors $$\begin{pmatrix}-\frac23&0&1&0&0\end{pmatrix}$$, $$\begin{pmatrix}0&-1&0&1&0\end{pmatrix}$$, and $$\begin{pmatrix}1&2&0&0&1\end{pmatrix}$$, these three vectors therefore span $$W$$

I have totally no idea how did they determine the parametrizing variables and got the three vectors.

$$W = \{(x_1,x_2,x_3,x_4,x_5) \in \Bbb{R}^5 : x_1+\frac{2}{3}x_3-x_5=0, x_2+x_4-2x_5=0\} \\ = \{(x_5 - \frac{2}{3}x_3,2x_5 -x_4,x_3,x_4,x_5) : x_3,x_4,x_5 \in \Bbb R\} \\ = \{x_3(-\frac{2}{3},0,1,0,0) + x_4(0,-1,0,1,0) + x_5(1,2,0,0,1) : x_3,x_4,x_5 \in \Bbb R\}.$$
Thats why the three vectors $$(-\frac{2}{3},0,1,0,0),(0,-1,0,1,0), (1,2,0,0,1)$$, spans $$W$$.
• "I have totally no idea how did they determine the parametrizing variables...": The non-parametrizing variables are the 1st and 2nd one because the 1st and 2nd columns contain a pivot. As a result, when you solve the system from the last line to the first, those variables will be expressed as linear combinations of the other ones: $$x_2=-x_4+2x_5,\quad x_1=-\frac23x_3+x_5,$$ hence so will be the whole generic solution: $$(x_1,\dots,x_5)=\left(-\frac23x_3+x_5,-x_4+2x_5,x_3,x_4,x_5\right).$$
• "...and got the three vectors." The three vectors $$u:=\begin{pmatrix}-\frac23&0&1&0&0\end{pmatrix},v:=\begin{pmatrix}0&-1&0&1&0\end{pmatrix},w:=\begin{pmatrix}1&2&0&0&1\end{pmatrix}$$ were obtained by setting $$(x_3,x_4,x_5)$$ to (respectively) $$(1,0,0),(0,1,0),(0,0,1),$$ and you can check that the generic solution is then equal to $$x_3u+x_4v+x_5w$$.